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Section 24.4 Multiplication

Subsection 24.4.1 Some coincidences

One surprising thing about end of the previous subsection is that the Euler products for the Riemann \(\zeta\) function and the Dirichlet series of the Möbius function are multiplicative inverses of each other. That is,

\begin{equation*} \prod_p \frac{1}{1-p^{-s}}=1/\left(\prod_p 1-p^{-s}\right)\text{.} \end{equation*}

We can check this numerically as well; in the following examples, we use \(s=2\text{.}\)

They agree up to quite a few digits when we approximate both representations of the number, so that is a start at reasonability!

Finally, recall from our exploration of the average value of \(\sigma\) in Section 20.4 that \(\zeta(2)=\frac{\pi^2}{6}\) (though there we just used this as a sum, and didn't call it \(\zeta(2)\)). Compare this computation with the ones above.

Remark 24.4.1.

Zeta has interesting values at integers, not just for \(s=2\text{.}\) Euler calculated many even values of \(\zeta\text{,}\) which all look like \(\pi^{2n}\) times a rational number (see any description of the so-called Bernoulli numbers). However, it was only in 1978 that \(\zeta(3)\) was shown to be irrational. It was then named Apéry's constant after the man who proved this, Roger Apéry.

To compare with the situation for even \(n\text{,}\) as of this writing it is still only known that at least one of the next four odd values (\(\zeta(5),\zeta(7),\zeta(9),\zeta(11)\)) is irrational 1 . See Wadim Zudilin's website for many links, though this page hasn't been updated for some time.

And various other similar facts.

Subsection 24.4.2 Multiplication of both kinds

Let's reinterpret this just a little bit. Assuming we can prove that all this makes sense (which we haven't, yet), we have the following two analogous facts.

This analogy is not a coincidence.

First, we need there is a key fact you may or may not have seen in calculus, related to absolute convergence (see for example Active Calculus). Roughly speaking, when series converge absolutely, you can mess around with them with a lot with impunity. See, for instance, Mertens' Theorem on convergence of Cauchy products. Interestingly, neither [E.4.6] nor [E.2.1, Theorem 9.6] say much more about this in their presentation of this standard proof. See Exercise 24.7.3 if you have not encountered this!

In any case, since \(F\) and \(G\) do converge absolutely, we can and will mess around a lot with the product

\begin{equation*} F(s)G(s)=\sum_{n=1}^\infty\frac{f(n)}{n^s}\sum_{m=1}^\infty\frac{g(m)}{m^s}\text{.} \end{equation*}

In particular, we can group the products by the terms \(\frac{f(n)g(m)}{n^sm^s}\) (the same way we did in proving things about \(\star\) in Subsection 23.4.3), without loss of equality.

We can further group by when \(n\) and \(m\) are complementary divisors of the same number (I suggest using specific numbers to try this out). This gives

\begin{equation*} F(s)G(s)=\sum_{d=1}^\infty\sum_{nm=d}\frac{f(m)g(n)}{d^s}\text{.} \end{equation*}

Notice that the inner sum is precisely the Dirichlet \(\star\) product (except divided by \(d^s\)). So we may rewrite this as

\begin{equation*} F(s)G(s)=\sum_{d=1}^\infty \frac{(f\star g)(d)}{d^s}\text{.} \end{equation*}

The numerators are the definition of \(h\text{,}\) so this is just \(H(s)\text{,}\) as desired. (In [E.4.6, Theorem 11.5] the additional detail that any Dirichlet series with these values must be the one for \(f\star g\) is proved, which requires a uniqueness result for the series we will omit.)

This is a quite remarkable and deep connection between the discrete/algebraic point of view and the analytic/calculus point of view. It is a shame that this is not exploited more in the standard calculus curriculum, though see [E.6.8] for a very good resource for those who wish to do so.